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#4*lnx=ln2x#

2 Answers
VNVDVI
Feb 25, 2018

#x=root(3)2#

Explanation:

First, recall that #aln(x)=ln(x^a)#, as per the exponent property for logarithms.

This means that #4*ln(x)=ln(x^4)#.

#ln(x^4)=ln(2x)#

Subtract #ln(2x)# from both sides, isolating all terms with natural logarithms:

#ln(x^4)-ln(2x)=0#

Recall that #ln(a)-ln(b)=ln(a/b)#, as per the quotient property for logarithms.

This means that #ln(x^4)-ln(2x)=ln(x^4/(2x))#

#ln(x^4/(2x))=0#

Simplify the term inside the natural logarithm:

#ln(x^(cancel(4)3)/(2cancelx))=0#

#ln(x^3/2)=0#

Exponentiate both sides with #e:#

#e^ln(x^3/2)=e^0#

#e^0=1# as anything to the power of #0# (with the exception of #0# itself) is equal to #1#.

#e^ln(x^3/2)=1#

Recall that #e^ln(x)=x# (proof shown below). This means that #e^ln(x^3/2)=x^3/2#

Solve for #x:#

#e^ln(x^3/2)=1#

#x^3/2=1#

#x^3=2#

#x=root(3)2#

Proof for #e^ln(x)=x#:

#e^ln(x)=x#

Apply the natural logarithm to both sides:

#ln(e^ln(x))=ln(x)#

#ln(e^ln(x))=ln(x)ln(e)#, exponent property for logarithms.

#ln(e)=log_ee=1#, as #log_aa=1#.

#ln(x)ln(e)=ln(x)#
#ln(x)(1)=ln(x)#
#ln(x)=ln(x)#
#x=x#

Douglas K.
Feb 25, 2018

Given: #4ln(x) = ln(2x)#

Add the implied domain restriction:

#4ln(x) = ln(2x); x > 0#

Use the property of logarithms #(c)ln(a) = ln(a^c)#:

#ln(x^4) = ln(2x); x > 0#

Raise both sides to the base:

#x^4= 2x; x > 0#

Write in standard form:

#x^4-2x = 0; x > 0#

Factor:

#x(x^3-2) = 0; x > 0#

We must discard the root #x = 0#

#x = root(3)2#

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